Answer
(a) The required torque is $~48.1~N \cdot m$
(b) Each child must push with a force of $19.2~N$
Work Step by Step
(a) Let $M_m$ be the mass of the merry-go-round and let $M_c$ be the mass of each child. We can find the rotational inertia of the system:
$I = \frac{1}{2}M_mR^2+2M_cR^2$
$I = \frac{1}{2}(350.0~kg)(1.25~m)^2+(2)(30.0~kg)(1.25~m)^2$
$I = 367.2~kg~m^2$
We can express the final angular speed in units of rad/s:
$25~rpm \times \frac{2\pi~rad}{1~rev} \times \frac{1~min}{60~s} = 2.618~rad/s$
We can find the angular acceleration:
$\omega_f = \omega_0+\alpha~t$
$\alpha = \frac{\omega_f - \omega_0}{t}$
$\alpha = \frac{2.618~rad/s-0}{20.0~s}$
$\alpha = 0.1309~rad/s^2$
We can find the required torque:
$\tau = I~\alpha$
$\tau = (367.2~kg~m^2)(0.1309~rad/s^2)$
$\tau = 48.1~N \cdot m$
The required torque is $~48.1~N \cdot m$
(b) We can find the force $F$ with which each child must push:
$2~(R \times F) = \tau$
$F = \frac{\tau}{2R}$
$F = \frac{48.1~N \cdot m}{(2)(1.25~m)}$
$F = 19.2~N$
Each child must push with a force of $19.2~N$.
